Question

Using the data provided, calculate the multiple bond energy $\left( kJ\, mol ^{-1}\right.$ ) of a $C \equiv C$ bond $C _{2} H _{2}$. That energy is (take the bond energy of a $C - H$ bond as $350 \,kJ \,mol ^{-1}$ )
$2 C (s)+ H _{2}(g) \longrightarrow C _{2} H _{2}(g) ; \Delta H=225\, kJ \,mol ^{-1}$
$2 C (s) \longrightarrow 2 C (g) ; \Delta H=1410 \,kJ \,mol ^{-1} $
$ H _{2}(g) \longrightarrow 2 H (g) ; \Delta H=330 \,kJ \,mol ^{-1} $

• A. 1165
• B. 837
• C. 865
• D. 815

Answer 1

Answer: (D)

For calculation of $C = C$ bond energy, we must first calculate dissociation energy of $C _{2} H _{2}$ as
$C _{2} H _{2}(g) \longrightarrow 2 C (g)+2 H (g)\,\,\,$...(i)
Using the given bond energies and enthalpies :
$C _{2} H _{2}(g) \longrightarrow 2 C (g)+2 H (g) ; \Delta H=-225 \,kJ\,\,\,$...(ii)
$2 C (s) \longrightarrow 2 C (g) ; \Delta H=1410 kJ \,\,\,$...(iii)
$H _{2}(g) \longrightarrow 2 H (g) ; \Delta H=330 kJ\,\,\, $...(iv)
Adding Eqs. (ii), (iii) and (iv) gives Eq. (i).
$\Rightarrow C _{2} H _{2}(g) \longrightarrow 2 C (g)+2 H (g) ; \Delta H=1515\, kJ$
$\Rightarrow 1515\, kJ =2 \times( C - H ) BE +( C \equiv C ) BE$
$=2 \times 350+( C \equiv C ) BE$
$\Rightarrow ( C \equiv C ) BE =1515-700=815 \,kJ / mol$